note-math

polar-decomposition-of-Lorentz-group_(tag) polar decomposition $SO (1, 3)$ to rotation and boost (ref-2, Vol.1, p.165)

let $𝐴 = (\begin{matrix} 𝜏 & 𝑣^{⊺} \\ 𝑢 & 𝑎 \end{matrix}) \in SO (1, 3)$ where $𝜏 \in ℝ, 𝑢, 𝑣 \in ℝ^{1, 3}, 𝑎 \in Matrix (3, ℝ)$

\begin{matrix} 𝐴 & = & (\begin{matrix} 1 \\ 𝑅 \end{matrix}) (\begin{matrix} {(1 + 𝑣^{⊺} 𝑣)}^{\frac{1}{2}} & 𝑣^{⊺} \\ 𝑣 & {(𝟙 + 𝑣 𝑣^{⊺})}^{\frac{1}{2}} \end{matrix}) \end{matrix}

(使用 $𝐴 \in SO (1, 3) ⟹ 𝐴^{⊺} 𝜂 𝐴 = 𝜂 = 𝐴 𝜂 𝐴^{⊺}$ 和 polar decomposition of positive definite symmetric matrix)

where

\begin{matrix} 𝑎 & = & 𝑅 {(𝑎^{⊺} 𝑎)}^{\frac{1}{2}} \\ 𝑎^{⊺} 𝑎 & = & 𝟙 + 𝑣 𝑣^{⊺} \\ 𝜏^{2} & = & 1 + 𝑣^{⊺} 𝑣 = 1 + 𝑢^{⊺} 𝑢 \\ 𝑎 𝑣 & = & 𝜏 𝑢 \end{matrix}

$𝐵 (𝑣) = (\begin{matrix} {(1 + 𝑣^{⊺} 𝑣)}^{\frac{1}{2}} & 𝑣^{⊺} \\ 𝑣 & {(𝟙 + 𝑣 𝑣^{⊺})}^{\frac{1}{2}} \end{matrix})$ is boost, map $(1, 0)$ to $(𝜏, 𝑣)$

$𝜏 = {(1 + 𝑣^{⊺} 𝑣)}^{\frac{1}{2}}$ , $𝟙 + 𝑣 𝑣^{⊺} = 𝑎^{⊺} 𝑎$ have diagonal form $(1, \dots, 1, 𝑒^{𝛼}, 𝑒^{- 𝛼})$ where $𝛼 = {cosh}^{- 1} 𝜏$

$𝐵 (𝑣) = (\begin{matrix} {(1 + 𝑣^{⊺} 𝑣)}^{\frac{1}{2}} & 𝑣^{⊺} \\ 𝑣 & {(𝟙 + 𝑣 𝑣^{⊺})}^{\frac{1}{2}} \end{matrix}) = exp (\begin{matrix} 𝑏^{⊺} \\ 𝑏 \end{matrix})$ with $\frac{sinh | 𝑏 |}{| 𝑏 |} 𝑏 = 𝑣, 𝜏 = cosh 𝑏$

Euler-angle-Lorentz-group_(tag) Question

rotation

使用 $𝑥_{1}, 𝑥_{2}$ 轴的旋转来生成 $SO (3)$

in $SO (3)$ , $𝜃 \in [0, 𝜋], 𝜃_{1}, 𝜃_{2} \in [0, 2 𝜋]$

\begin{matrix} 𝑅 (𝜃, 𝜃_{1}, 𝜃_{2}) \\ = & (\begin{matrix} 1 \\ cos 𝜃_{1} & - sin 𝜃_{1} \\ sin 𝜃_{1} & cos 𝜃_{1} \end{matrix}) \cdot (\begin{matrix} cos 𝜃 & - sin 𝜃 \\ 1 \\ sin 𝜃 & cos 𝜃 \end{matrix}) \cdot (\begin{matrix} 1 \\ cos 𝜃_{2} & - sin 𝜃_{2} \\ sin 𝜃_{2} & cos 𝜃_{2} \end{matrix}) \end{matrix}

in $SU (2)$

\begin{matrix} 𝑅 (𝜃, 𝜃_{1}, 𝜃_{2}) & = & (\begin{matrix} 𝑒^{\frac{1}{2} 𝜃_{1} i} \\ 𝑒^{- \frac{1}{2} 𝜃_{1} i} \end{matrix}) \cdot (\begin{matrix} cos \frac{1}{2} 𝜃 & i sin \frac{1}{2} 𝜃 \\ i sin \frac{1}{2} 𝜃 & cos \frac{1}{2} 𝜃 \end{matrix}) \cdot (\begin{matrix} 𝑒^{\frac{1}{2} 𝜃_{2} i} \\ 𝑒^{- \frac{1}{2} 𝜃_{2} i} \end{matrix}) \\ = & (\begin{matrix} cos \frac{1}{2} 𝜃 \cdot 𝑒^{\frac{1}{2} (𝜃_{1} + 𝜃_{2}) i} & i sin \frac{1}{2} 𝜃 \cdot 𝑒^{\frac{1}{2} (𝜃_{1} - 𝜃_{2}) i} \\ i sin \frac{1}{2} 𝜃 \cdot 𝑒^{- \frac{1}{2} (𝜃_{1} - 𝜃_{2}) i} & cos \frac{1}{2} 𝜃 \cdot 𝑒^{- \frac{1}{2} (𝜃_{1} + 𝜃_{2}) i} \end{matrix}) \end{matrix}

boost

使用 $𝑥_{1}, 𝑥_{2}$ 轴的 boost

in $SO (1, 3)$

\begin{matrix} 𝐵 (𝜑, 𝜑_{1}, 𝜑_{2}) \\ = & (\begin{matrix} cosh 𝜑_{1} & sinh 𝜑_{1} \\ sinh 𝜑_{1} & cosh 𝜑_{1} \\ 1 \\ 1 \end{matrix}) \cdot (\begin{matrix} cosh 𝜑 & sinh 𝜑 \\ 1 \\ sinh 𝜑 & cosh 𝜑 \\ 1 \end{matrix}) \cdot (\begin{matrix} cosh 𝜑_{2} & sinh 𝜑_{2} \\ sinh 𝜑_{2} & cosh 𝜑_{2} \\ 1 \\ 1 \end{matrix}) \end{matrix}

in $SL (2, ℂ)$

\begin{matrix} 𝐵 (𝜑, 𝜑_{1}, 𝜑_{2}) & = & (\begin{matrix} 𝑒^{\frac{1}{2} 𝜑_{1}} \\ 𝑒^{- \frac{1}{2} 𝜑_{1}} \end{matrix}) \cdot (\begin{matrix} cosh \frac{1}{2} 𝜑 & sinh \frac{1}{2} 𝜑 \\ sinh \frac{1}{2} 𝜑 & cosh \frac{1}{2} 𝜑 \end{matrix}) \cdot (\begin{matrix} 𝑒^{\frac{1}{2} 𝜑_{2}} \\ 𝑒^{- \frac{1}{2} 𝜑_{2}} \end{matrix}) \\ = & (\begin{matrix} cosh \frac{1}{2} 𝜑 \cdot 𝑒^{\frac{1}{2} (𝜑_{1} + 𝜑_{2})} & sinh \frac{1}{2} 𝜑 \cdot 𝑒^{\frac{1}{2} (𝜑_{1} - 𝜑_{2})} \\ sinh \frac{1}{2} 𝜑 \cdot 𝑒^{- \frac{1}{2} (𝜑_{1} - 𝜑_{2})} & cosh \frac{1}{2} 𝜑 \cdot 𝑒^{- \frac{1}{2} (𝜑_{1} + 𝜑_{2})} \end{matrix}) \end{matrix}

Lorentz-group-Lie-bracket_(tag) $so (1, 3)$ with boost and rotation decomposition $𝑏 + 𝑟 = (𝜑_{1} 𝑏_{1} + 𝜑_{2} 𝑏_{2} + 𝜑_{3} 𝑏_{3}) + (𝜃_{1} 𝑟_{1} + 𝜃_{2} 𝑟_{2} + 𝜃_{3} 𝑟_{3})$ and Lie-bracket

\begin{matrix} 𝑏 \times 𝑏 & = & - 𝑟 \\ 𝑏 \times 𝑟 & = & 𝑏 \\ 𝑟 \times 𝑏 & = & 𝑏 \\ 𝑟 \times 𝑟 & = & 𝑟 \\ 𝑏 \cdot 𝑟 & = & 𝑟 \cdot 𝑏 \end{matrix}

其中 $\times$ 模仿 $ℝ^{3}$ cross product. Example $𝑏 \times 𝑏 = (\begin{matrix} [𝑏_{2}, 𝑏_{3}] \\ [𝑏_{3}, 𝑏_{1}] \\ [𝑏_{1}, 𝑏_{2}] \end{matrix})$

其中 $\cdot$ 模仿 $ℝ^{3}$ dot product. Example $𝑏 \cdot 𝑟 = 𝑟 \cdot 𝑏$ ==> $𝑏_{1} 𝑟_{1} = 𝑟_{1} 𝑏_{1}$ or $[𝑏_{1}, 𝑟_{1}] = 0$

写为模仿 $ℂ^{3}$ cross product

\begin{matrix} \frac{1}{2} (𝑟 + 𝑏 i) \times \frac{1}{2} (𝑟 + 𝑏 i) & = & \frac{1}{2} i (𝑟 + 𝑏 i) \\ (𝑟 + 𝑏 i) \times (𝑟 - 𝑏 i) & = & 0 \\ (𝑟 + 𝑏 i) \cdot (𝑟 - 𝑏 i) & = & 𝑟^{2} + 𝑏^{2} \end{matrix}

$so (1, 3)$ have form $(\begin{matrix} 0 & 𝑏^{⊺} \\ 𝑏 & 𝑟 \end{matrix})$ where $𝑟 \in so (3), 𝑏 \in ℝ^{3}$ (ref-2, Vol.1, p.180)

Lorentz-group-orbit-isotropy_(tag)

$SO (1, 3)$ or $SL (2, ℂ)$ act on $ℝ^{1, 3}$

orbit type	isotropy-group type
$\| 𝑥 \|^{2} = 1$	$SO (3)$ or $SU (2, ℂ)$
$\| 𝑥 \|^{2} = - 1$	$SO (1, 2)$ or $SL (2, ℝ)$
$\| 𝑥 \|^{2} = 0$	$SO (2) ⋊ ℝ^{2}$
${0}$	$SO (1, 3)$ or $SL (2, ℂ)$

isotropy-on-lightcone_(tag) Prop $SO (1, 3)$ 作用于 lightcone 类似于 $SO (2) ⋊ ℝ^{2}$ (恰好是 $ℝ^{2}$ Euclidean 仿射群)

Proof 使用 #link(<spacetime-momentum-aciton-spinor-representation>)[spinor 技术]

Prop $SL (2, ℂ)$ 作用在 lightcone (不是 projective-lightcone), #link(<isotropy>)[] 类似于 $U (1) ⋊ ℂ$

$SL (2, ℂ)$ 是 #link(<action-surjective>)[满射作用], orbit 数 $1$ , 所以计算 isotropy #link(<isotropy-in-same-orbit-is-isom>)[只需要考虑] 一点

使用光锥射影上的点 $𝑝 = (\begin{matrix} 1 \\ 1 \\ 0 \\ 0 \end{matrix}) \leftrightarrow 𝑝_{spin} = (\begin{matrix} 2 \end{matrix}) \in ℝ^{1, 3}$ , 计算 isotropy $𝐴 𝑝_{spin} 𝐴^{†} = 𝑝_{spin}$ , where $𝐴 𝑝_{spin} 𝐴^{†} = (\begin{matrix} 2 𝑎 𝑎^{*} & 2 𝑎 𝑐^{*} \\ 2 𝑎^{*} 𝑐 & 2 𝑐 𝑐^{*} \end{matrix})$

$𝑐 = 0, | 𝑎 |^{2} = 1$ 是光锥 isotropy
$𝑐 = 0, | 𝑎 |^{2} \in ℝ$ 是对 $(\begin{matrix} 1 \\ 1 \\ 0 \\ 0 \end{matrix})$ 的伸缩
$SU (2)$ 是对 $(\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) \in ℝ^{3}$ 的空间旋转. 能给出整个光锥截面 $𝕊^{2}$

==> isotropy $𝐴 = (\begin{matrix} 𝑒^{𝜃} & 𝑏 \\ 𝑒^{- 𝜃} \end{matrix})$ where $𝜃 \in Im (ℂ)$

类似于 #link(<isotropy-on-projective-lightcone>)[] 的计算, 此处将类似于 $U (1) ⋊ ℂ$

isotropy-on-lightcone-intuition_(tag) isotropy-group of orbit lightcone 的直观. 根据

(\begin{matrix} 𝑒^{\frac{1}{2} 𝜃 i} & 0 \\ 𝑒^{- \frac{1}{2} 𝜃 i} \end{matrix}) (\begin{matrix} 1 & 𝑏 \\ 1 \end{matrix}) (\begin{matrix} 𝑒^{\frac{1}{2} 𝜃 i} & 0 \\ 𝑒^{- \frac{1}{2} 𝜃 i} \end{matrix}) = (\begin{matrix} 𝑒^{𝜃 i} & 𝑏 \\ 𝑒^{- 𝜃 i} \end{matrix})

分开讨论两种情况

$(\begin{matrix} 𝑒^{\frac{1}{2} 𝜃 i} & 0 \\ 𝑒^{- \frac{1}{2} 𝜃 i} \end{matrix})$ . is rotation in $𝑝_{1}$
$(\begin{matrix} 1 & 𝑏 \\ 1 \end{matrix})$

let $so (1, 3)$ with boost and rotation decomposition $𝑏 + 𝑟 = (𝜑_{1} 𝑏_{1} + 𝜑_{2} 𝑏_{2} + 𝜑_{3} 𝑏_{3}) + (𝜃_{1} 𝑟_{1} + 𝜃_{2} 𝑟_{2} + 𝜃_{3} 𝑟_{3})$ (not the $𝑏$ in $(\begin{matrix} 𝑒^{𝜃 i} & 𝑏 \\ 𝑒^{- 𝜃 i} \end{matrix})$ )

线性同构到新的基

\begin{matrix} 𝑏_{1}, 𝑟_{1} \\ 𝑏_{2} + 𝑟_{3}, 𝑏_{3} + 𝑟_{2} \\ 𝑏_{2} - 𝑟_{3}, 𝑏_{3} - 𝑟_{2} \end{matrix}

where

$𝑟_{3}$ is rotation in $𝑝_{1}, 𝑝_{2}$
$𝑏_{2}$ is boost in $𝑝_{0}, 𝑝_{2}$
$𝑏_{2} - 𝑟_{3}$ and $𝑏_{3} - 𝑟_{2}$ 是 lightcone coordinate $𝑝_{0} \pm 𝑝_{1}$ 类似物, 保持 $(\begin{matrix} 1 \\ 1 \\ 0 \\ 0 \end{matrix})$

或者写为

\begin{matrix} 𝑏_{1}, 𝑟_{1} \\ 𝑀_{+} \\ 𝑀_{-} \end{matrix}

where $𝑏_{1}, 𝑀_{+}$ 将会改变 $(\begin{matrix} 1 \\ 1 \\ 0 \\ 0 \end{matrix})$ , $𝑟_{1}, 𝑀_{-}$ 则固定

二维 Lie algebra $𝑀_{-}$ is commutative, 对应 $U (1) ⋊ ℂ$ 中的 $𝑏 \in ℂ$ or $(\begin{matrix} 1 & 𝑏 \\ 1 \end{matrix})$

在光锥上 (图)

一般 $𝑏$ 生成双曲型 orbit
一般 $𝑟$ 生成椭圆型 orbit
$𝑀_{+}$ 可以推广到一般的 lightcone 组合 e.g. $𝑝_{0} \pm 𝑝_{2}$ , 将生成抛物型 orbit

$(\begin{matrix} 1 & 𝑏 \\ 1 \end{matrix})$ 作用的具体计算

let $𝑧 = 𝑝_{0} + 𝑝_{1}, 𝑧^{*^{'}} = 𝑝_{0} - 𝑝_{1}, 𝑤 = 𝑝_{2} + 𝑝_{3} i, 𝑤^{*} = 𝑝_{2} - 𝑝_{3} i$ . metric will be $| 𝑧 |^{' 2} + | 𝑤 |^{2} = 𝑧 𝑧^{*^{'}} + 𝑤 𝑤^{*}$

\begin{matrix} (\begin{matrix} 1 & 𝑏 \\ 1 \end{matrix}) (\begin{matrix} 𝑧 & 𝑤 \\ 𝑤^{*} & 𝑧^{*^{'}} \end{matrix}) (\begin{matrix} 1 \\ 𝑏^{*} & 1 \end{matrix}) \\ = & (\begin{matrix} 𝑧 + | 𝑏 |^{2} 𝑧^{*^{'}} + 2 Re (𝑏 𝑤) & 𝑤 + 𝑏 𝑧^{*^{'}} \\ 𝑤^{*} + 𝑏^{*} 𝑧^{*^{'}} & 𝑧^{*^{'}} \end{matrix}) \end{matrix}

in $ℝ^{1, 3}$

(\begin{matrix} 𝑝_{0} + \frac{1}{2} | 𝑏 |^{2} (𝑝_{0} - 𝑝_{1}) + Re (𝑏 (𝑝_{2} + 𝑝_{3} i)) \\ 𝑝_{1} + \frac{1}{2} | 𝑏 |^{2} (𝑝_{0} - 𝑝_{1}) + Re (𝑏 (𝑝_{2} + 𝑝_{3} i)) \\ 𝑝_{2} + Re (𝑏 (𝑝_{0} - 𝑝_{1})) \\ 𝑝_{3} + Im (𝑏 (𝑝_{0} - 𝑝_{1})) \end{matrix})